Answer:
The initial number of bacteria is 371.
Explanation:
Let P represents population of the bacteria and t represents time,
According to the question,
[tex]\frac{dP}{dt}\propto P[/tex]
[tex]\implies \frac{dP}{dt}=kP[/tex]
Where, k is constant of proportionality,
[tex]\frac{dP}{dt}=kdt[/tex]
Integrating both sides,
[tex]ln P=kt+C[/tex]
[tex]P=e^{kt+C}[/tex]
[tex]P=e^C e^{kt}[/tex]
[tex]P=P_0 e^{kt}[/tex] ( Let [tex]P_0=e^C[/tex] )
If t = 0, [tex]P=P_0[/tex]
That is, [tex]P_0[/tex] is the intial population.
Since, when t = 3, P = 500 and when t = 10, P=4000
[tex]500=P_0 e^{3t}[/tex]
[tex]4000 = P_0 e^{10t}[/tex]
[tex]\because \frac{4000}{500}=\frac{ P_0 e^{10t}}{ P_0 e^{3t}}[/tex]
[tex]8=e^{(10-3)t}=e^{7t}\implies 7t = ln(8) \implies t=\frac{ln(8)}{7}=0.297063077383\approx 0.2971[/tex]
[tex]\implies 500=P_0 e^{0.2971}\implies P_0=\frac{500}{e^{0.2971}}= 371.484855838\approx 371[/tex]
Hence, initial population is approximately 371.
The initial number of bacteria was 7 bacterial cells.
P (t) = P 0 and rt
Where:
P(t) = the amount of some quantity at time t
P 0 = initial amount at time t = 0
r = growth rate
t = time (number of periods)
Initial amount (P 0 ) 7.21940555138e-32
Growth rate (r) 8 = 800.0%
Time (t) 10
Final amount (P (t)) 4000
With this information, we can conclude that the initial rounded number of bacterial cells is 7.
